Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x - 4} = \dfrac{6x - 6}{x - 4}$
Explanation: Multiply both sides by $x - 4$ $ \dfrac{x^2 + 2}{x - 4} (x - 4) = \dfrac{6x - 6}{x - 4} (x - 4)$ $ x^2 + 2 = 6x - 6$ Subtract $6x - 6$ from both sides: $ x^2 + 2 - (6x - 6) = 6x - 6 - (6x - 6)$ $ x^2 + 2 - 6x + 6 = 0$ $ x^2 + 8 - 6x = 0$ Factor the expression: $ (x - 4)(x - 2) = 0$ Therefore $x = 4$ or $x = 2$ At $x = 4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 4$, it is an extraneous solution.